Question: Simplify and expand the following expression: $ \dfrac{4}{q - 3}- \dfrac{1}{4q + 40}- \dfrac{4}{q^2 + 7q - 30} $
Answer: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $4$ out of denominator in the second term: $ \dfrac{1}{4q + 40} = \dfrac{1}{4(q + 10)}$ We can factor the quadratic in the third term: $ \dfrac{4}{q^2 + 7q - 30} = \dfrac{4}{(q - 3)(q + 10)}$ Now we have: $ \dfrac{4}{q - 3}- \dfrac{1}{4(q + 10)}- \dfrac{4}{(q - 3)(q + 10)} $ The least common multiple of the denominators is: $ (q - 3)(q + 10)$ In order to get the first term over $(q - 3)(q + 10)$ , multiply by $\dfrac{4(q + 10)}{4(q + 10)}$ $ \dfrac{4}{q - 3} \times \dfrac{4(q + 10)}{4(q + 10)} = \dfrac{16(q + 10)}{(q - 3)(q + 10)} $ In order to get the second term over $(q - 3)(q + 10)$ , multiply by $\dfrac{q - 3}{q - 3}$ $ \dfrac{1}{4(q + 10)} \times \dfrac{q - 3}{q - 3} = \dfrac{q - 3}{(q - 3)(q + 10)} $ In order to get the third term over $(q - 3)(q + 10)$ , multiply by $\dfrac{4}{4}$ $ \dfrac{4}{(q - 3)(q + 10)} \times \dfrac{4}{4} = \dfrac{16}{(q - 3)(q + 10)} $ Now we have: $ \dfrac{16(q + 10)}{(q - 3)(q + 10)} - \dfrac{q - 3}{(q - 3)(q + 10)} - \dfrac{16}{(q - 3)(q + 10)} $ $ = \dfrac{ 16(q + 10) - (q - 3) - 16} {(q - 3)(q + 10)} $ Expand: $ = \dfrac{16q + 160 - q + 3 - 16}{4q^2 + 28q - 120} $ $ = \dfrac{15q + 147}{4q^2 + 28q - 120}$